## Elementary Differential Equations and Boundary Value Problems 9th Edition

$$4 y^{\prime \prime}-4 y^{\prime}-3 y=0$$ The general solution of that equation is given by $$y=c_{1} e^{\frac{3}{2}t} +c_{2} e^{ -\frac{1}{2}t}$$ where $c_{1}$ and $c_{2}$ are arbitrary constants.
$$4 y^{\prime \prime}-4 y^{\prime}-3 y=0 \quad \quad (1)$$ We assume that $y = e^{rt}$, and it then follows that $r$ must be a root of the characteristic equation $$4r^{2}-4 r-3=(r+\frac{1}{3})^{2}=(r-\frac{3}{2})(r+\frac{1}{2})=0,$$ Thus the possible values of $r$ are $r_{1}=\frac{3}{2} , r_{2}=-\frac{1}{2}$; the general solution of Eq. (1) is $$y=c_{1} e^{\frac{3}{2}t} +c_{2} e^{ -\frac{1}{2}t}$$ where $c_{1}$ and $c_{2}$ are arbitrary constants.